Lie Derivative for Wedge Product of Vector Fields I am having trouble here. The context is: Let $X$, $Y$ and $S$ be vector fields ina a manifold (we can assume it's $\mathbb{C}^2$ though I'm pretty sure this should work in any manifold), and we can work fine with the exterior product or wedge product of vector fields, since they are tensors.
I need to know if the formula
$$
\mathcal{L}_X(S\wedge Y)=\mathcal{L}_X(S)\wedge Y+S\wedge\mathcal{L}_X(Y)
$$
I know this is true when $X$, $Y$ and $S$ are differential forms. The demonstration is basd solely on the property that says that, for tensors fields, we have
$$
\mathcal{L}_X(S\otimes Y)=\mathcal{L}_X(S)\otimes Y+S\otimes\mathcal{L}_X(Y)
$$
I don't think I can say that since it is correct for the tensor product, it would be for the exterior product. I guess I must use the fact that vector fields are antissimetric 1-linear forms and use the operator (in my reference it is called "anti-simetrization operator")
$$
\mathcal{\alpha}(X)=\sum_{s\in \mathcal{G}_p} \epsilon (s)s\circ X
$$
where $\mathcal{G}_p$ is the set or permitations of $p$ indexes and the composition means a permutation on the indexes of the base elements of $X$. The application $\alpha$ turns linear p-forms into antissimetric forms and then we have exterior product of those. If $X$ is already antissimetric, then $\alpha(X)=p!X $.
Now, we also have the definition
$$
X\wedge Y=\dfrac{1}{p!q!}\alpha (X\otimes Y)
$$
So I'm guessing I can argue that $\alpha (X\otimes Y)=(p+q)!(X\otimes Y)$, and the calculations work, that is, I get the expression $\mathcal{L}_X(S\wedge Y)=\mathcal{L}_X(S)\wedge Y+S\wedge\mathcal{L}_X(Y)$ as I wanted. But I don't know for sure if this is correct. I am trying to self-learn somethings on tensors. 
Can someone tell me if it's correct and, if not, point me my mistakes? 
$$
$$
 A: Let us consider the following general setting. We need it to prove the statement. 


*

*The dg Lie algebra of poly-vector fields
Let $M$ be a real manifold of dimension $n$ over the ground field $\mathbb K$. Let 
$$\operatorname{T}^{\bullet}_{poly}(M):=\mathcal C^{\infty}(M)\otimes_{\mathbb K}\wedge^{\bullet+1}\operatorname{T}(M) $$
be the algebra of poly vector fields on $M$. Note the shift of grading: for example, vector fields are polyvectors of degree $0$.
There exists a structure of differential graded Lie algebra on $\operatorname{T}^{\bullet}_{poly}(M)$ given as follows. The differential is equal to $0$. The Lie bracket is the Schouten bracket $[\cdot,\cdot]_\mathcal{S} $ given by
$$[e_1 ∧ ... ∧ e_k, \eta_1 ∧ ... ∧ \eta_l]_\mathcal{S} =
\\
\sum_{i=1}^k\sum_{j=1}^l(−1)^{i+j}\mathcal L_{e_i}(\eta_j)\wedge e_1 \wedge\dots\wedge\hat{e}_i\wedge\dots\wedge e_k\wedge \eta_1\wedge\dots\wedge\hat{\eta}_l\wedge\dots\wedge\eta_l, $$
for all $e_{\bullet}$ and $\eta_{\bullet}$ in $\operatorname{T}^{0}_{poly}(M)$ and denoting omission by $\hat{\cdot}$. Note that the Schouten bracket reduces to the Lie bracket
$$\mathcal L_X(Y):=[X,Y],$$
on $\operatorname{T}^{0}_{poly}(M)$. 
In summary, using some lengthy but straightforward computations, one can prove that
$$(\operatorname{T}^{\bullet}_{poly}(M),0,[\cdot,\cdot]_\mathcal{S}) $$
is a dg Lie algebra (I do not want to introduce the exact definition and further discuss the gradings). Note that we have also an associative product, i.e. the wedge product. In other words, the structure on $\operatorname{T}^{\bullet}_{poly}(M)$ is even richer, but let us skip the discussion about Gerstenhaber algebras.


*

*Statement in the OP
In the above setting, the original statement is equivalent to

For all $X,Y,S\in \operatorname{T}^{0}_{poly}(M)$ the  identity
$$[X,S\wedge Y]_\mathcal{S}=[X,S]\wedge Y+S\wedge[X,Y]~~(*) $$
holds.

On the l.h.s. of $(*)$ it is necessary to consider the Schouten bracket because $S\wedge Y\in \operatorname{T}^{1}_{poly}(M)$.
Let us prove it; by definition of the Schouten bracket
$$[X,S\wedge Y]_\mathcal{S}=(−1)^{1+1}\mathcal L_{X}(S)\wedge Y+(−1)^{1+2}\mathcal L_{X}(Y)\wedge S=\mathcal L_{X}(S)\wedge Y-\mathcal L_{X}(Y)\wedge S,  $$
or
$$[X,S\wedge Y]_\mathcal{S}=\mathcal L_{X}(S)\wedge Y+S\wedge\mathcal L_{X}(Y),  $$
as claimed. This ends the proof.
A: @Marra: Avirus gave a nice explanation of the Schouten-Nijenhuis bracket (useful in Poisson geometry). If you want just to know the Lie derivative of the exterior product
$\mathscr{L}_X(Y\wedge Z)$, then you can start from the definition, for any tensor field $T$:
$$\mathscr{L}_X T=\frac{d}{dt}\Big|_{t=0}(\exp tX)^*T$$
where $\exp tX$ the local flow of $X$. From there you can prove, for any tensor fields $S$ and $T$, that 
$$\mathscr{L}_X(S\wedge T)=\mathscr{L}_X S\wedge T+S\wedge\mathscr{L}_X T.$$
Finally, it's useful to keep in mind that a Lie derivative is always a derivation:
$$\mathscr{L}_X(fg)=\mathscr{L}_X(f)g+f\mathscr{L}_X(g),\ \forall f,g\in C^\infty(M).$$
$$\mathscr{L}_X(\alpha\wedge\beta)=\mathscr{L}_X\alpha\wedge\beta+\alpha\wedge\mathscr{L}_X\beta,\ \forall\alpha,\beta\in\Omega^*(M).$$
$$\mathscr{L}_X<\alpha,Y>=<\mathscr{L}_X\alpha,Y>+<\alpha,\mathscr{L}_X Y>,\ \forall\alpha,\in\Omega^1(M),\ \forall Y\in\mathcal{X}(M).$$
and the general definition of a derivation: if $(\mathcal{A},\cdot)$ is an algebra, a derivation on $\mathcal{A}$ is linear map $\delta : \mathcal{A}\to\mathcal{A}$ such that,
$$\delta(x\cdot y)=(\delta x)\cdot y+x\cdot(\delta y).$$
P.S: Sorry for my bad English!
